# Find the probability mass function of Y=sum_{i=1}^{N} X_i where X_i∼Geometric(a) and N∼Geometric(b). I attempted to do this by finding the probability generating function of Y and comparing it to known probability generating functions to take advantage of the uniqueness property. (In my searches online, it sounds like I should find that Y∼Geometric(ab).)

Geometric sum of geometric random variables
I am looking to find the probability mass function of $Y=\sum _{i=1}^{N}{X}_{i}$ where ${X}_{i}\sim \text{Geometric}\left(a\right)$ and $N\sim \text{Geometric}\left(b\right)$. I attempted to do this by finding the probability generating function of Y and comparing it to known probability generating functions to take advantage of the uniqueness property. (In my searches online, it sounds like I should find that $Y\sim \text{Geometric}\left(ab\right)$.)
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Cody Petty
Step 1
Assume that $\begin{array}{}\text{(1)}& \mathbb{P}\left[{X}_{i}=k\right]=p\left(1-p{\right)}^{k-1},\phantom{\rule{2em}{0ex}}k=1,2,\dots \end{array}$
and $\mathbb{P}\left[N=n\right]=r\left(1-r{\right)}^{n-1}$ for $r=1,2,\dots$. Then: $\begin{array}{rcl}\mathbb{P}\left[Y=m\right]& =& \sum _{n\ge 1}\mathbb{P}\left[N=n\right]\cdot \mathbb{P}\left[{X}_{1}+\dots +{X}_{n}=m\right]\\ \text{(2)}& & =& \frac{r}{1-r}\sum _{n\ge 1}\left(1-r{\right)}^{n}{\left(\frac{p}{1-p}\right)}^{n}\left(1-p{\right)}^{m}\cdot r\left(m,n\right)\end{array}$
Where: $\begin{array}{rcl}r\left(m,n\right)& =& \mathrm{#}\left\{\left({a}_{1},\dots ,{a}_{n}\right):{a}_{i}\in {\mathbb{N}}_{>0},{a}_{1}+\dots +{a}_{n}=m\right\}\\ & =& \left[{x}^{m}\right]{\left(x+{x}^{2}+{x}^{3}+\dots \right)}^{n}\\ \text{(3)}& & =& \left[{x}^{m}\right]\frac{{x}^{n}}{\left(1-x{\right)}^{n}}=\left(\genfrac{}{}{0}{}{m-1}{n-1}\right)\end{array}$
Step 2
Hence: $\begin{array}{rcl}\mathbb{P}\left[Y=m\right]& =& rp\left(1-p{\right)}^{m-1}\sum _{n\ge 1}\left(\genfrac{}{}{0}{}{m-1}{n-1}\right){\left(\frac{p-pr}{1-p}\right)}^{n-1}\\ & =& rp\left(1-p{\right)}^{m-1}{\left(1+\frac{p-pr}{1-p}\right)}^{m-1}\\ \text{(4)}& & =& pr\left(1-pr{\right)}^{m-1}\end{array}$
proving your claim for geometric distributions supported on 1,2,…. There is little to change in order to deal with geometric distributions supported on 0,1,…, too.